As given that speed of car as well as truck
$$
=72 \mathrm{~km} / \mathrm{h}=72 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}
$$
For retarded motion of truck,
$$
v=u+a_t t \quad\left(\because v=0, a_t=?, t=5 \mathrm{sec} .\right)
$$
$$
0=20+a_t \times 5
$$
or $a_t=-4 \mathrm{~m} / \mathrm{s}^2$
$t=3 \mathrm{sec}, u=20 \mathrm{~m} / \mathrm{sec}, v=0, a_c=$ ?
For retarded motion of the car,
$$
\begin{aligned}
&v=u+a_c t \\
&0=20+a_c \times 3
\end{aligned}
$$
So, $a_c=-\frac{20}{3} \mathrm{~m} / \mathrm{s}^2$
Let car be at a distance $x$ meter from the truck, when truck gives the signal and $t$ be the time taken to cover this distance.
The time of human response is $0.5 \mathrm{~s}$, time of retarded motion of car is $(t-0.5) \mathrm{s}$.
Velocity of car after time $t$,
$$
\begin{aligned}
v_c &=u+a_c t \\
&=20-\left(\frac{20}{3}\right)(t-0.5)
\end{aligned}
$$
Velocity of truck after time $t$
$$
v_t=20-4 t
$$
To avoid the car bump onto the truck,
$$
\begin{aligned}
&v_c=v_t \\
&20-\frac{20}{3}(t-0.5)=20-4 t \\
&\Rightarrow 4 t=\frac{20}{3}(t-0.5) \Rightarrow 3 t=5 t-2.5 \Rightarrow t=1.25 \text { seconds }
\end{aligned}
$$
Distance travelled by car and truck in time $t=\frac{5}{4} \sec$.
$$
\begin{aligned}
&s_t=u_t t+\frac{1}{2} a_t t^2 \\
&=20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2=25-3.125 \\
&\therefore s_t=21.875 \mathrm{~m}
\end{aligned}
$$
Distance travelled by car in time $t$
$=$ Distance travelled by car in $0.5 \mathrm{~s}$ (without retardation) $+$ Distance travelled by car in $(t-0.5) \mathrm{s}$ (with retardation)
$$
\begin{aligned}
&s_c=(20 \times 0.5)+20\left(\frac{5}{4}-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(\frac{5}{4}-0.5\right)^2 \\
&s_c=23.125 \mathrm{~m} \\
&s_c-s_t=23.125-21.875=1.250 \mathrm{~m}
\end{aligned}
$$
So avoid the bump onto the truck, the car must maintain a distance from the truck atleast $1.250 \mathrm{~m}$.